A285305 Fixed point of the morphism 0 -> 10, 1 -> 1001.

1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0

Offset

1

Comments

This is a 3-automatic sequence. See Allouche et al. link. - Michel Dekking, Oct 05 2020

Links

Clark Kimberling, Table of n, a(n) for n = 1..10000

J.-P. Allouche, F. M. Dekking, and M. Queffélec, Hidden automatic sequences, arXiv:2010.00920 [math.NT], 2020.

Index entries for sequences that are fixed points of mappings

Formula

Conjecture: a(n) = A284775(n+1). - R. J. Mathar, May 08 2017

From Michel Dekking, Feb 20 2021: (Start)

Proof of this conjecture. Let mu: 0 -> 10, 1 -> 1001 be the defining morphism of (a(n)), and let tau: 0 -> 01, 1 -> 0011 be the defining morphism of A284775.

Since mu^n(0) starts with 1 for n>1, mu^n(0) tends to (a(n)) as n tends to infinity. So the conjecture will follow directly from the following claim.

CLAIM: 0 mu^n(0) = tau^n(0) 0 for n>0.

Proof: By induction over two levels, exploiting the obvious equality tau(1) = 0 tau(0) 1 to go from the third to the fourth line below.

For n=1: 0 mu(0)= 010 = tau(0) 0.

For n=2: 0 mu^2(0)= 0100110 = tau^2(0) 0.

Suppose true for n-1 and n. Then

tau^{n+1}(0) =

tau^n(tau(0)) =

tau^n(0) tau^n(1) =

tau^n(0) tau^{n-1)(0) tau^n(0) tau^{n-1}(1) =

0 mu^n(0)0^{-1} 0 mu^{n-1}(0)0^{-1}0mu^n(0)0^{-1}tau^{n-1)(1)=

0 mu^{n-1}(mu(0))mu^{n-1}(0)mu^{n-1}(mu(0))0^{-1}0tau^{n-)(1)=

0 mu^{n-1}(10) mu^{n-1}(0) mu^{n-1}(10) 0^{-1} tau^{n-1)(1) =

0 mu^{n-1}(1001) mu^{n-1}(0) 0^{-1} tau^{n-1)(1) =

0 mu^n(1) 0^{-1} tau^{n-1)(0) 0 0^{-1} tau^{n-1)(1) =

0 mu^n(1) 0^{-1} tau^{n-1}(01) =

0 mu^n(1) 0^{-1} tau^n(0) =

0 mu^n(1) mu^n(0) 0^{-1} =

0 mu^n(mu(0)) 0^{-1} =

0 mu^{n+1}(0) 0^{-1}.

So we proved 0 mu^{n+1}(0) = tau^{n+1}(0) 0.

(End)

Example

0 -> 10-> 1001 -> 100110101001 ->

Mathematica

s = Nest[Flatten[# /. {0 -> {1, 0}, 1 -> {1, 0, 0, 1}}] &, {0}, 10]; (* A285305 *)
u = Flatten[Position[s, 0]];  (* A285306 *)
v = Flatten[Position[s, 1]];  (* A285307 *)

Crossrefs

Cf. A284306, A285307, A284775.

Partial sums of A284793.

Sequence in context: A285258 A068428 A078650 * A028863 A089012 A083035

Adjacent sequences: A285302 A285303 A285304 * A285306 A285307 A285308

Keyword

nonn,easy

Author

Clark Kimberling, Apr 25 2017

Status

approved